It is said that if the product of $n$, the number of trials, and $p$, the probability of a success is large, then a binomial distribution can be accurately approximated by a normal distribution.
Is the theory supporting this the Central Limit Theorem? When I think of central limit theorems, I usually think of the sum or mean of a series of IID random variables, where the sum or mean approaches a normal distribution as the number of variables approaches infinity. However, in the current case, I don't see sums or means, so is the idea that the binomial distribution can be approximately as normal because of CLT or some other condition?
Elaborating on the document cited in the OP's comment, the claim is, that the hypotheses $0\le p_n,q_n\le 1$, $np_n\to\infty$, and $ nq_n\to\infty$ (where $q_n=1-p_n$) together imply that the CLT applies to $X_n\sim \operatorname{Bin}(n,p_n)$, in the sense that $$\lim_{n\to\infty}P\left(\frac{X_n-np_n}{\sqrt{np_nq_n}}<x\right)= \lim_{n\to\infty}F_n(x) = \Phi(x)$$ for all $x$. (Where $F_n$ is the cdf of the standardized version of $X_n$, and $\Phi$ is the standard normal cdf.) In the same vein, one could ask if the same conclusion followed under the simpler-looking hypothesis that $np_nq_n\to\infty$.
This version differs from the original statement of the problem by imposing a condition on $n q_n$, as does the web page cited in the OP's comment.
By the Berry-Essen theorem (a sharpening of the usual central limit theorem) we know that there is a constant $C$ such that for all $n$, and real $x$, $$ |F_n(x)-\Phi(x)|\le \frac {C\rho_n}{\sigma_n^3\sqrt n}=B_n\text{ say},$$ where $C$ is a constant, $\sigma_n=\sqrt{p_nq_n}$, and $\rho_n=p_nq_n(p_n^2+q_n^2)$. Note that if $Z_n=-p_n$ with probabilty $q_n$ and $Z_n=1-p_n$ with probability $p_n$, then $E[Z_n]=0$, $\sigma_n^2=E[Z_n^2]$, and $\rho_n=E[|Z_n|^3]$.
So now one just notices that $$B_n=\frac{C(p_n^2+q_n^2)}{\sqrt{np_nq_n}}=\frac{Cp_n^{3/2}}{\sqrt{n q_n}} + \frac{Cq_n^{3/2}}{\sqrt{n p_n}} = O\left(\frac 1{\sqrt{np_n}} + \frac 1{\sqrt {nq_n}}\right) = o(1),$$ under the first set of hypoptheses. If one assumes $np_nq_n\to\infty$ the result follows similarly: $B_n\le C/\sqrt{np_nq_n}=o(1)$. In either case, if you want $B_n$ to be less than (say) $1/10$, this tells you what your thresholds on $np_n$ and $nq_n$ must be, and so on.